Difference between revisions of "2005 AMC 10A Problems/Problem 25"
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Therefore, the desired ratio is <math>\frac{133\sin A}{392\sin A}=\frac{19}{56}\Longrightarrow \mathrm{(D)}</math> | Therefore, the desired ratio is <math>\frac{133\sin A}{392\sin A}=\frac{19}{56}\Longrightarrow \mathrm{(D)}</math> | ||
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| + | Note: <math>BC=39</math> was not used in this problem | ||
==See also== | ==See also== | ||
Revision as of 11:07, 8 November 2015
Problem
In 
we have 
, 
, and 
. Points 
and 
are on 
and 
respectively, with 
and 
. What is the ratio of the area of triangle 
to the area of the quadrilateral 
?
Solution
.
Using this formula:
![$[ADE]=\frac{1}{2}\cdot19\cdot14\cdot\sin A = 133\sin A$](http://latex.artofproblemsolving.com/4/9/9/4994450bce1d1f4bb79c88144f39f85e954923c3.png)
![$[ABC]=\frac{1}{2}\cdot25\cdot42\cdot\sin A = 525\sin A$](http://latex.artofproblemsolving.com/9/d/5/9d54255e8dd4f8c76b2fe7fb3a02cc6a1a7bd0dc.png)
Since the area of is equal to the area of minus the area of ,
![$[BCED] = 525\sin A - 133\sin A = 392\sin A$](http://latex.artofproblemsolving.com/2/3/8/238b4f47585fadb1facd70b51395692d972ba99d.png)
.
Therefore, the desired ratio is 
Note: 
was not used in this problem
See also
| 2005 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
